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Exam ASTAM · Buhlmann and Buhlmann-Straub · Tutorial 532 of 979

Bias Correction in VHM Estimation

Learn why sample variation among risk means overstates VHM and how to subtract the appropriate process-variance contribution for balanced and unequal exposures.

Intermediate 9 min read

What You'll Learn

  • Distinguish VHM from process variance in observed risk means
  • Derive the expected process-variance contribution to an unweighted sample variance
  • Apply the correction when every risk has the same exposure
  • Calculate the correction for unequal exposures using unweighted risk means
  • Derive and apply the correction for an exposure-weighted dispersion estimator
  • Recognize when a corrected VHM estimate is negative and interpret the result

Why Observed Risk Means Need a Correction

In the Bühlmann-Straub model, observed experience for each risk reflects two sources of variation. Risks differ in their underlying expected outcomes, and each risk's observed experience also fluctuates around its own expected outcome. Consequently, the spread of observed risk means is generally larger than the spread of the underlying risk means alone. Treating the entire observed spread as VHM (variance of hypothetical means) therefore overstates VHM.

The correction depends on how risk means are summarized and on the exposures supporting them. With equal exposures, the process contribution has a simple form. With unequal exposures, a risk mean based on less exposure has greater process variance, so the correction must account for the exposure pattern. This tutorial derives both the unweighted and exposure-weighted corrections.

Definition: Let \(r\) risks have latent means \(\mu_i\), where the \(\mu_i\) are independent draws from a population with mean \(m\) and variance \(\operatorname{Var}(\mu_i)=\text{VHM}\). Let \(X_{ij}\) be the observation for risk \(i\) in period \(j\), and let \(w_{ij}\) be its exposure. Under the Bühlmann-Straub assumptions, observations are conditionally independent given their risk means, and \(\operatorname{E}[X_{ij}\mid\mu_i]=\mu_i\) and \(\operatorname{Var}(X_{ij}\mid\mu_i)=\text{EPV}/w_{ij}\), where EPV is the expected process variance per exposure unit. For risk \(i\), define total exposure \(w_i=\sum_j w_{ij}\) and experience mean \(\bar X_i=\sum_j w_{ij}X_{ij}/w_i\).

The exposure-scaled variance assumption means that more exposure reduces the process variance of a risk's experience mean. Conditional independence gives \(\operatorname{Var}(\bar X_i\mid\mu_i)=\text{EPV}/w_i\). The observed risk mean thus contains both its latent risk mean and sampling fluctuation around that mean.

Unweighted Sample Variance: Derivation of the Correction

An unweighted estimator gives every risk mean equal influence, regardless of its exposure. Define the ordinary sample mean of the risk means as \(\bar X_{\cdot}=r^{-1}\sum_i\bar X_i\), and their sample variance as \(S^2_{\bar X}=(r-1)^{-1}\sum_i(\bar X_i-\bar X_{\cdot})^2\). The same risk-level variance decomposition applies whether exposures are equal or unequal; the exposure pattern determines the process contribution.

$$ \operatorname{E}[S^2_{\bar X}] =\text{VHM}+\frac{1}{r}\sum_{i=1}^{r}\frac{\text{EPV}}{w_i} =\text{VHM}+\frac{\text{EPV}}{r}\sum_{i=1}^{r}\frac{1}{w_i}. $$

To see why, write \(\bar X_i=\mu_i+\varepsilon_i\), where \(\operatorname{E}[\varepsilon_i\mid\mu_i]=0\) and \(\operatorname{Var}(\varepsilon_i)=\text{EPV}/w_i\). The risk means are independent across risks and have common expectation \(m\), but their variances are \(\text{VHM}+\text{EPV}/w_i\). For independent observations with common expectation, the expected ordinary sample variance is the average of their variances. Specifically, if their variances are \(\sigma_i^2\), then \(\operatorname{E}[\sum_i(\bar X_i-\bar X_{\cdot})^2]=\sum_i\sigma_i^2-r\operatorname{Var}(\bar X_{\cdot})\), and \(\operatorname{Var}(\bar X_{\cdot})=r^{-2}\sum_i\sigma_i^2\). Dividing by \(r-1\) gives \(r^{-1}\sum_i\sigma_i^2\). Substituting \(\sigma_i^2=\text{VHM}+\text{EPV}/w_i\) yields the result.

Result: The process-variance correction for the unweighted sample variance is \(\text{EPV}\,r^{-1}\sum_i 1/w_i\). Thus the unweighted method-of-moments estimate is \(\widehat{\text{VHM}}=S^2_{\bar X}-\widehat{\text{EPV}}\,r^{-1}\sum_i 1/w_i\), where \(\widehat{\text{EPV}}\) is an estimate of EPV. If every risk has exposure \(w\), this simplifies to subtracting \(\widehat{\text{EPV}}/w\).

Worked Example: Correcting a Balanced Portfolio

A hypothetical portfolio has four risks, each with total exposure \(w_i=3\). Their experience means are \(7,9,11,\) and \(13\). An independent within-risk analysis estimates EPV as 12. Estimate VHM using the unweighted sample variance and its process correction.

Method. Calculate the sample variance of the four risk means, then subtract EPV divided by the common exposure. The common trap is subtracting EPV itself, rather than the process variance of an exposure-weighted mean.

The mean of the risk means is \((7+9+11+13)/4=10\). The sum of squared deviations is \((7-10)^2+(9-10)^2+(11-10)^2+(13-10)^2=9+1+1+9=20\). Therefore \(S^2_{\bar X}=20/(4-1)=20/3\).

The correction is \(12/3=4\). Hence \(\widehat{\text{VHM}}=20/3-4=20/3-12/3=8/3=2.6667\). The observed sample variance, \(6.6667\), exceeds the corrected estimate by exactly 4, the process-variance contribution. Values are rounded to four decimal places.

Check. With equal exposures, the formula requires the same correction for each risk mean, \(12/3=4\). The corrected estimate is positive and the subtraction is consistent with the observed risk means containing process noise.

Unequal Exposures: Two Estimators, Two Corrections

When exposures differ, there are two natural ways to summarize the spread of risk means. An unweighted sample variance gives each risk equal weight and uses the average reciprocal exposure correction derived above. An exposure-weighted dispersion gives more influence to more-exposed risks and has a different denominator and correction. These estimators should not be mixed: the correction must match the dispersion statistic used.

Result: For the exposure-weighted method, let \(W=\sum_i w_i\), \(\bar X_w=\sum_i w_i\bar X_i/W\), and \(D=W-\sum_i w_i^2/W\). The weighted dispersion is \(Q_w=\sum_i w_i(\bar X_i-\bar X_w)^2\). Under the model assumptions, \(\operatorname{E}[Q_w]=\text{VHM}\,D+(r-1)\text{EPV}\). Therefore the corrected estimate is \(\widehat{\text{VHM}}=[Q_w-(r-1)\widehat{\text{EPV}}]/D\), provided \(D>0\).

For justification, write the weighted dispersion in quadratic form as \(\sum_i w_i\bar X_i^2-(\sum_i w_i\bar X_i)^2/W\). The \(\text{VHM}\) part of its expectation is \(\text{VHM}(W-\sum_iw_i^2/W)=\text{VHM}\,D\). The process part is \(\sum_iw_i(\text{EPV}/w_i)-W^{-1}\sum_iw_i^2(\text{EPV}/w_i)=r\text{EPV}-\text{EPV}=(r-1)\text{EPV}\). Subtracting that contribution and dividing by \(D\) removes the expected process variation while putting the remaining dispersion on the VHM scale. If exposures are all \(w\), then \(D=(r-1)w\), and the correction to \(Q_w/D\) reduces to \(\text{EPV}/w\), agreeing with the balanced result.

Worked Example: Weighted Correction With Unequal Exposures

A hypothetical insurer observes three risk means: \(10,12,\) and \(14\), with exposures \(1,2,\) and \(4\), respectively. An independent analysis gives EPV \(=6\). Calculate the exposure-weighted corrected estimate of VHM.

Method. Use \(Q_w\) and the weighted correction \((r-1)\text{EPV}/D\). A common mistake is to use the unweighted correction formula with a weighted dispersion.

First, \(W=1+2+4=7\). The weighted mean is \(\bar X_w=(1(10)+2(12)+4(14))/7=90/7\). The denominator is \(D=7-(1^2+2^2+4^2)/7=7-21/7=4\).

Compute the weighted dispersion using the equivalent sum-of-squares expression: \(Q_w=1(10^2)+2(12^2)+4(14^2)-(90^2/7)=1172-8100/7=(8204-8100)/7=104/7\). Thus \(Q_w/D=(104/7)/4=26/7\).

The process correction on the VHM scale is \((3-1)6/4=12/4=3\). Therefore \(\widehat{\text{VHM}}=26/7-3=(26-21)/7=5/7=0.7143\). Rounding is to four decimal places.

Check. The corrected dispersion is \(5/7\), and adding back the correction gives \(5/7+3=26/7\), the uncorrected weighted dispersion on the VHM scale.

Worked Example: Unequal Exposures With an Unweighted Statistic

A hypothetical book has three risk means \(8,10,\) and \(12\), based on exposures \(1,2,\) and \(4\). An independent estimate of EPV is 6. Estimate VHM using the ordinary, unweighted sample variance of the risk means.

Method. Use the unweighted estimator, not the exposure-weighted one. Its correction is EPV times the average reciprocal exposure.

The risk-mean average is \((8+10+12)/3=10\), so \(S^2_{\bar X}=[(8-10)^2+(10-10)^2+(12-10)^2]/2=(4+0+4)/2=4\).

The average reciprocal exposure is \((1+1/2+1/4)/3=(7/4)/3=7/12\). The process correction is \(6(7/12)=42/12=3.5\). Therefore the raw corrected estimate is \(4-3.5=0.5\). The final estimate is 0.5, with no rounding needed.

Check. The least-exposed risk has the largest process variance, 6, while the exposure-4 risk has process variance \(6/4=1.5\). The correction, 3.5, is their average process variance, as required for an unweighted sample variance.

Interpreting the Estimate and Avoiding Exam Traps

These corrections remove expected process variation; they do not guarantee that every finite-sample estimate will be positive. A corrected method-of-moments estimate can be negative if the observed dispersion is smaller than the estimated process contribution. That result indicates insufficient observed between-risk variation to support a positive raw estimate under this calculation. If a procedure imposes a nonnegative constraint and replaces a negative estimate by zero, that constrained value is not the same unbiased method-of-moments estimate.

The main exam trap is using a correction that belongs to a different statistic. For the unweighted sample variance, subtract \(\text{EPV}\) times the average of \(1/w_i\). For the weighted dispersion divided by \(D\), subtract \((r-1)\text{EPV}/D\). In balanced data, both give \(\text{EPV}/w\); this agreement is a useful check, but does not make the unequal-exposure formulas interchangeable.

1
Identify the statistic.
Decide whether the observed spread is an ordinary sample variance of risk means or a weighted dispersion.
2
Calculate the matching process term.
For an unweighted variance use \(\widehat{\text{EPV}}\,r^{-1}\sum_i1/w_i\). For a weighted dispersion use \((r-1)\widehat{\text{EPV}}/D\) on the VHM scale.
3
Subtract and interpret.
Keep the exposure units consistent, show the raw corrected value, and state if it is negative rather than silently changing the estimator.
Key takeaway: Observed variation among risk means combines VHM and process variation. Correct VHM estimation requires subtracting the process contribution implied by both the exposure pattern and the exact dispersion statistic being used.

Check Your Understanding

For each question, identify the estimator before applying its correction. Answers are given so that the calculations can be checked.

  1. Five risks have equal exposure \(w_i=2\), EPV is 8, and their unweighted sample variance is 9. Find the corrected VHM estimate. (Answer: 5)
  2. Four risks have exposures \(1,2,2,4\), EPV is 12, and the unweighted sample variance of their means is 10. Find the correction and corrected estimate. (Answer: correction \(=6.75\); estimate \(=3.25\))
  3. Three risks have exposures \(1,1,2\), risk means \(5,9,7\), and EPV is 4. Using the weighted estimator, calculate \(D\), \(Q_w\), and the corrected VHM estimate. (Answer: \(D=2.5\), \(Q_w=8\), estimate \(=0\))
  4. For an unweighted sample variance with exposures \(2,3,6\) and EPV 9, calculate the process-variance correction. (Answer: 3)
  5. Four risks have exposures \(1,2,3,4\) and EPV 5. For the exposure-weighted estimator, calculate \(D\) and the process correction on the VHM scale. (Answer: \(D=7\); correction \(=15/7\approx2.1429\))
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